Due in 20 minutes!! please help

Expain the diffents between numerical expressions and algebraic expressions

Oduvanchick [21]

<h2>Numerical expressions are numbers and signs, like below. The following are some examples of numerical expressions, numerical expressions do not use letters. 4 + 20 – 7, (2 + 3) – 7, (6 × 2) ÷ 20, 5 ÷ (20 × 3) An algebraic expression uses letters and it says you to find "x/y/z". Example; 4+20xy-7x, (2x+3) - 7y+20.</h2>

7 0

3 years ago

Factor n^3 - n^2 + 3n - 3.

lozanna [386]

N^2(n - 1) + 3(n - 1)

(n^2 + 3)(n - 1)

the answer is c

5 0

3 years ago

What is the distance around a triangle that has sides measuring 2 1/8, 3 1/2, and 2 1/2 feet?

Mashcka [7]

Perimeter = 2 1/8 + 3 1/2 + 2 1/2 = 7 (1 + 4 + 4)/8 = 7 9/8 = 8 1/8

3 0

3 years ago

The airplane Li’s family will be flying on can seat up to 149 passengers. If 96 passengers are currently on the plane, which ine

Ivenika [448]

For this case, the first thing you should do is define a variable.
We have then:
x: number of passengers remaining who can board the plane.
We have as data:
1) They can board up to 149 passengers
2) There are 96 passengers currently aboard.
Writing inequality we have:
$x + 96 \leq 149
$
Answer:
An inequality that can be used to determine how many more people can board is:
$x + 96 \leq 149$

3 0

3 years ago

Read 2 more answers

A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

netineya [11]

Answer:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=19, p=0.3)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 5)$

And we can use the complement rule:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

4 0

3 years ago